NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2#

We factorise the polynomial:

$$x^2 - 2x - 8 = x^2 - 4x + 2x - 8$$

$$= x(x - 4) + 2(x - 4)$$

$$= (x - 4)(x + 2)$$

The zeroes are obtained by setting each factor to zero:

$$x - 4 = 0 \implies x = 4$$

$$x + 2 = 0 \implies x = -2$$

So the zeroes are $\alpha = 4$ and $\beta = -2$.

Verification:

Here $a = 1$, $b = -2$, $c = -8$.

$\alpha + \beta = 4 + (-2) = 2 = -\dfrac{(-2)}{1} = -\dfrac{b}{a}$ ✓

$\alpha\beta = 4 \times (-2) = -8 = \dfrac{-8}{1} = \dfrac{c}{a}$ ✓

We factorise:

$$4s^2 - 4s + 1 = 4s^2 - 2s - 2s + 1$$

$$= 2s(2s - 1) - 1(2s - 1)$$

$$= (2s - 1)(2s - 1)$$

$$= (2s - 1)^2$$

Setting to zero: $2s - 1 = 0 \implies s = \dfrac{1}{2}$ (repeated root)

So both zeroes are $\alpha = \beta = \dfrac{1}{2}$.

Verification:

Here $a = 4$, $b = -4$, $c = 1$.

$\alpha + \beta = \dfrac{1}{2} + \dfrac{1}{2} = 1 = -\dfrac{(-4)}{4} = -\dfrac{b}{a}$ ✓

$\alpha\beta = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} = \dfrac{c}{a}$ ✓

Rewriting: $6x^2 - 7x - 3$

We factorise:

$$6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3$$

$$= 3x(2x - 3) + 1(2x - 3)$$

$$= (2x - 3)(3x + 1)$$

Zeroes: $2x - 3 = 0 \implies x = \dfrac{3}{2}$ and $3x + 1 = 0 \implies x = -\dfrac{1}{3}$

So $\alpha = \dfrac{3}{2}$, $\beta = -\dfrac{1}{3}$.

Verification:

Here $a = 6$, $b = -7$, $c = -3$.

$\alpha + \beta = \dfrac{3}{2} - \dfrac{1}{3} = \dfrac{9 - 2}{6} = \dfrac{7}{6} = -\dfrac{(-7)}{6} = -\dfrac{b}{a}$ ✓

$\alpha\beta = \dfrac{3}{2} \times \left(-\dfrac{1}{3}\right) = -\dfrac{1}{2} = \dfrac{-3}{6} = \dfrac{c}{a}$ ✓

$$4u^2 + 8u = 4u(u + 2)$$

Zeroes: $4u = 0 \implies u = 0$ and $u + 2 = 0 \implies u = -2$

So $\alpha = 0$, $\beta = -2$.

Verification:

Here $a = 4$, $b = 8$, $c = 0$.

$\alpha + \beta = 0 + (-2) = -2 = -\dfrac{8}{4} = -\dfrac{b}{a}$ ✓

$\alpha\beta = 0 \times (-2) = 0 = \dfrac{0}{4} = \dfrac{c}{a}$ ✓

$$t^2 - 15 = (t - \sqrt{15})(t + \sqrt{15})$$

Zeroes: $t = \sqrt{15}$ and $t = -\sqrt{15}$

So $\alpha = \sqrt{15}$, $\beta = -\sqrt{15}$.

Verification:

Here $a = 1$, $b = 0$, $c = -15$.

$\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0 = -\dfrac{0}{1} = -\dfrac{b}{a}$ ✓

$\alpha\beta = \sqrt{15} \times (-\sqrt{15}) = -15 = \dfrac{-15}{1} = \dfrac{c}{a}$ ✓

We factorise:

$$3x^2 - x - 4 = 3x^2 - 4x + 3x - 4$$

$$= x(3x - 4) + 1(3x - 4)$$

$$= (3x - 4)(x + 1)$$

Zeroes: $3x - 4 = 0 \implies x = \dfrac{4}{3}$ and $x + 1 = 0 \implies x = -1$

So $\alpha = \dfrac{4}{3}$, $\beta = -1$.

Verification:

Here $a = 3$, $b = -1$, $c = -4$.

$\alpha + \beta = \dfrac{4}{3} + (-1) = \dfrac{4 - 3}{3} = \dfrac{1}{3} = -\dfrac{(-1)}{3} = -\dfrac{b}{a}$ ✓

$\alpha\beta = \dfrac{4}{3} \times (-1) = -\dfrac{4}{3} = \dfrac{-4}{3} = \dfrac{c}{a}$ ✓

Sum of zeroes $= \alpha + \beta = \dfrac{1}{4}$

Product of zeroes $= \alpha\beta = -1$

The required quadratic polynomial is:

$$k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]$$

$$= k\left[x^2 - \dfrac{1}{4}x - 1\right]$$

Taking $k = 4$:

$$= 4x^2 - x - 4$$

∴ The quadratic polynomial is $\mathbf{4x^2 - x - 4}$.

Sum of zeroes $= \alpha + \beta = \sqrt{2}$

Product of zeroes $= \alpha\beta = \dfrac{1}{3}$

The required polynomial:

$$k\left[x^2 - \sqrt{2}\,x + \dfrac{1}{3}\right]$$

Taking $k = 3$:

$$= 3x^2 - 3\sqrt{2}\,x + 1$$

∴ The quadratic polynomial is $\mathbf{3x^2 - 3\sqrt{2}\,x + 1}$.

Sum of zeroes $= \alpha + \beta = 0$

Product of zeroes $= \alpha\beta = \sqrt{5}$

The required polynomial:

$$k\left[x^2 - 0 \cdot x + \sqrt{5}\right] = k\left[x^2 + \sqrt{5}\right]$$

Taking $k = 1$:

∴ The quadratic polynomial is $\mathbf{x^2 + \sqrt{5}}$.

Sum of zeroes $= \alpha + \beta = 1$

Product of zeroes $= \alpha\beta = 1$

The required polynomial:

$$k\left[x^2 - x + 1\right]$$

Taking $k = 1$:

∴ The quadratic polynomial is $\mathbf{x^2 - x + 1}$.

Sum of zeroes $= \alpha + \beta = -\dfrac{1}{4}$

Product of zeroes $= \alpha\beta = \dfrac{1}{4}$

The required polynomial:

$$k\left[x^2 + \dfrac{1}{4}x + \dfrac{1}{4}\right]$$

Taking $k = 4$:

$$= 4x^2 + x + 1$$

∴ The quadratic polynomial is $\mathbf{4x^2 + x + 1}$.

Sum of zeroes $= \alpha + \beta = 4$

Product of zeroes $= \alpha\beta = 1$

The required polynomial:

$$k\left[x^2 - 4x + 1\right]$$

Taking $k = 1$:

∴ The quadratic polynomial is $\mathbf{x^2 - 4x + 1}$.